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2 years ago

An object speeding up, an object slowing down, and an object changing direction are all examples of acceleration.

Physics

1 answer:

emmainna [20.7K]2 years ago

6 0

<h2>Answer: True</h2>

When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down.

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What is the value of 9682 when rounded to three significant figures

sdas [7]

9700 because 682 rounds to 700

5 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago

Which group of elements are shiny, opaque, and have a high melting point?

vlabodo [156]

The answer is A. Metals

3 0

3 years ago

What is newton's second law of motion?

lord [1]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

3 0

3 years ago

The wheels of an automobile are locked as it slides to a stop from an initial speed of 30.0 m/s. If the coefficient of kinetic f

Amiraneli [1.4K]

Answer:

x = 76.5 m

Explanation:

Let's use Newton's second law at the point of contact between the wheel and the floor.

fr = m a

fr = miy N

N-W = 0

N = W

μ mg = m a

a = miu g

a = 0.600 9.8

a = 5.88 m / s²

Having the acceleration we can use the kinematic relationships to find the distance

$v_{f}$ ² = v₀² + 2 a x

$v_{f}$ = 0

x = -v₀² / 2 a

Acceleration opposes the movement by which negative

x = - 30²/2 (-5.88)

x = 76.5 m

8 0

3 years ago